Proving Vector Norm Properties Using Different Definitions

[mathnerd15]

Homework Statement

this is an Apostol problem in chapter 12 and I guess it's a hypothetical definition of a norm of a vector
Assuming this different definition of the norm prove these statements-
$$Def. ||A||=\sum_{k=1}^{n}|a_{k}|, prove ||A||>0, if ||A||\neq0,||A||=0 if A=0, ||cA||=c||A||,triangle equality ||A+B||\leq ||A||+||B||,$$

I just looked at this- do I just expand the sums out and then express them in sigma notation?

Use this definition in V2 and prove on a figure the set of all points (x,y) of norm 1- this is just the line x+y=1?

which of the above theorems/statements would hold if we take the absolute value of the summation?

$$Def. ||A||=|\sum_{k=1}^{n}a_{k}|$$

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

Homework Statement

this is an Apostol problem in chapter 12 and I guess it's a hypothetical definition of a norm of a vector
Assuming this definition of the norm prove these statements-

$$Def. ||A||=\sum_{k=1}^{n}a_{k}\text{, prove }||A||>0\text{, if }||A||\neq0,$$

You haven't told us what $A$ is or what the $a_k$ are. If $n = 2$ is $v = \langle 1,-1\rangle \in A$? If so $v \ne 0$ but $\|v\| = 0$.

 

[mathnerd15]

Mostly I'm asking about the triangle inequality proof, the problem just gives you a general vector A
I think the hypothetical definition is ||A||=|a1|+|a2|+|a3|...+|an|, so if ||A|| norm doesn't equal the zero vector (0,0...0) and at least one component |ak|>0, then the sum must be >0

 

[LCKurtz]

Mostly I'm asking about the triangle inequality proof, the problem just gives you a general vector A
I think the hypothetical definition is ||A||=|a1|+|a2|+|a3|...+|an|, so if ||A|| norm doesn't equal the zero vector (0,0...0) and at least one component |ak|>0, then the sum must be >0

You "think" that is the norm definition? It's your question -- is it or isn't it? I see you edited your post to change it to that. And given your last sentence, what is your question?

[Edit] Re the triangle inequality, what have you tried?

 

[mathnerd15]

I'm just trying the TE for the first time. Apostol gives this 'incorrect' definition of the norm and asks you to prove the TE. I proved the law of cosines in norm form. here I see it is correct for a case where the signs of A and B are different

$$Def. ||A||=\sum_{k=1}^{n}|a_{k}|, triangle.inequality ||A+B||\leq ||A||+||B||. e.g.A=-2,B=1,|-1|\leq 3$$

 

[LCKurtz]

I'm just trying the TE for the first time. Apostol gives this 'incorrect' definition of the norm and asks you to prove the TE. I proved the law of cosines in norm form. here I see it is correct for a case where the signs of A and B are different

$$Def. ||A||=\sum_{k=1}^{n}|a_{k}|, triangle.inequality ||A+B||\leq ||A||+||B||. e.g.A=-2,B=1,|-1|\leq 3$$

What do you mean the signs of A and B? That doesn't make any sense. They are vectors, not scalars. You have to start by calculating A+B and its norm and compare it with the norms of A and B.

 

[mathnerd15]

I've attached the written problems, is this how I prove the triangle equality for this definition of norm, I ended up with something that seems analogous to Cauchy-Schwarz inequality?

anyways I started to lose confidence after thinking my handwriting was too messy after I saw a video of Andrew Wiles

 

[LCKurtz]

I've attached the written problems, is this how I prove the triangle equality for this definition of norm, I ended up with something that seems analogous to Cauchy-Schwarz inequality?

anyways I started to lose confidence after thinking my handwriting was too messy after I saw a video of Andrew Wiles

You are making it way too hard. Like I said before, calculate A+B and its norm. You don't need to square anything. Just use a familiar inequality. Show your work here, not in a pdf.

 

[mathnerd15]

$$||A+B||\leqslant ||A||+||B||\rightarrow \sum_{k=1}^{n}|ak+bk|\leq \sum_{k=1}^{n}|ak|+\sum_{k=1}^n|bk|$$
so this is trivially true by properties of absolute value? there are many more beautiful properties of absolute value functions

the last part of the solution is
|x|+|y|=1 and y=1-x, y=x+1, y=x-1, y=-x-1, and since |x|+|y|=1, then these lines become segments, 0<=x<=1 constrained and form a rectangle in V2 (similar to R2?)

 

[LCKurtz]

$$||A||+||B||\leqslant ||A+B||\rightarrow \sum_{k=1}^{n}|ak|+ \sum_{k=1}^{n}|bk|\leq \sum_{k=1}^{n}|ak+bk|$$
so this is trivially true by properties of absolute value?

No, that is so wrong it leaves me speechless. One more time I will ask you:

1. Write down A = ... and B = ... and then what is A + B is.
2. Then write down its norm $\|A+B\| =$?
Then you are ready to try to get the inequality.

 

[mathnerd15]

Sorry I'm not sure what you mean, the text gives you a hypothetical definition of norm as the sum of absolute values of |ak|
the true definition of norm as dot product gives a proof of the triangle equality from algebra or as a consequence of the Cauchy-Schwarz inequality

 

[LCKurtz]

No, that is so wrong it leaves me speechless. One more time I will ask you:

1. Write down A = ... and B = ... and then what is A + B is.
2. Then write down its norm $\|A+B\| =$?
Then you are ready to try to get the inequality.

Sorry I'm not sure what you mean, the text gives you a hypothetical definition of norm as the sum of absolute values of |ak|
the true definition of norm as dot product gives a proof of the triangle equality from algebra or as a consequence of the Cauchy-Schwarz inequality

Sorry, but I will not download a pdf file for a simple problem like this. You can easily type it. But until you do 1. and 2. above, I don't see where we have anything to discuss.

The Cauchy-Schwarz inequality has nothing to do with this problem.

[Edit] I now see you have changed post #9 to make it correct, without noting that it was changed. That makes it very difficult to maintain continuity of the thread and is against forum rules.

 

[mathnerd15]

$$A=\sum_{k=1}^{n}akEk, B=\sum_{k=1}^{n}bkEk, ||A||=\sqrt{A\cdot A}. \\ ||A+B||=\sum_{k=1}^{n}[(ak+bk)^{2}]^{1/2}, \\ but here ||A||=\sum_{k=1}^{n}|ak|, ||B||=\sum_{k=1}^{n}|bk|, \\ ||A||+||B||=\sum_{k=1}^{n}(|ak|+|bk|), ||A+B||=\sum_{k=1}^{n}\|ak+bk| \\ \sum_{k=1}^{n}\|ak+bk| <=\sum_{k=1}^{n}(|ak|+|bk|)$$


I'm sorry, so all vectors in Vn can be expressed as linear combinations of the orthogonal basis vectors [Ek,En] or other arbitrarily rotated orthogonal basis vectors as well as the imaginary basis vector on the C plane? I've written what I think is the definition of norm/length and also the hypothetical one (I'm not sure if there's some significance to Apostol's hypothetical problem)

 

[mathnerd15]

You haven't told us what $A$ is or what the $a_k$ are. If $n = 2$ is $v = \langle 1,-1\rangle \in A$? If so $v \ne 0$ but $\|v\| = 0$.

actually since ||A|| is defined as the sum of absolute values then if v=(1,-1) then ||v||=2, not 0, while normally
$$||v||=2^{1/2}$$
it is really the greatest achievement of humanity! by the way what would be an example of a difficult proof to try that is more advanced or a text?

 

[mathnerd15]

am I correct in the last post?